$\begin{aligned} y&=\cos^4(x) \\\\ \dfrac{dy}{dx}&=\,? \end{aligned}$ Choose 1 answer: Choose 1 answer: (Choice A) A $\cos(4x^3)$ (Choice B) B $-4\sin^3(x)\cos(x)$ (Choice C) C $-4x^3\sin(x^4)$ (Choice D) D $-4\cos^3(x)\sin(x)$
Since $\cos^4(x)$ is a composite function, we can use the chain rule. The chain rule says: $\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right]={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)}$ To use it, we need to recognize our function as a composite function like $w\big(u(x)\big)$. First, we realize that $\cos^4(x)$ is just shorthand for $(\cos(x))^4$, which is clearly composite: $\underbrace{(~\overbrace{\cos(x)}^{\text{inner}}~)^4}_{\text{outer}}$ So if $\cos^4(x)=w(u(x))$, then: $\begin{aligned} {u(x)}&={\cos(x)} &&\text{inner function} \\\\ w(x)&=x^4&&\text{outer function} \end{aligned}$ Before applying the chain rule, let's find the derivatives of the inner and outer functions (work not shown, but hopefully these derivatives are familiar by now). $\begin{aligned} {u'(x)}&={-\sin(x)} \\\\ {w'(x)}&={4x^3} \end{aligned}$ Now let's apply the chain rule: $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{d}{dx}\left[w\Bigl(u(x)\Bigr)\right] \\\\ &={w'\Bigl({u(x)}\Bigr)}\cdot{u'(x)} \\\\ &={4({\cos(x)})^3} \cdot {-\sin(x)} \\\\ &=-4\cos^3(x)\sin(x) \end{aligned}$